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Here is my proof.

Let U be an union of open sets and suppose x belongs to U.

Suppose there is largest open interval I that contains x.

Then Consider U-I, if this is empty, then we are done.

If U-I is not empty, pick any point y in U-I and let J be the largest open interval that contains y. In general, intersection of J and I must be empty, otherwise this would contradict our assumption.

If we keep doing this process, we would end up getting disjoint Union of open sets.

Is this rigorous proof?

Andrés E. Caicedo
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jessie
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  • It’s not rigorous: to make it so, you’d have to prove that for each $x\in U$ there is a largest open interval $I$ such that $x\in I\subseteq U$. Proving that is really the heart of the argument. That’s a pretty big hole that isn’t trivial to fix, so I’m going to close this as a duplicate: at the earlier question you'll find a variety of proofs. – Brian M. Scott Oct 27 '15 at 22:33

1 Answers1

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U is the disjoint union of its connected components. Each connected component of U is an interval whose interior is an interval since each element of U is contained in an interval.

Tsemo Aristide
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